An air-cored solenoid with length 30 ; cm,are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the soleno

Vedclass · Accepted Answer

$6.5$

Vedclass · Answer

(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the solenoid,$N = 500$Current in the solenoid,$I = 2.5 \; A$Time interval,$\Delta t = 10^{-3} \; s$The magnetic field inside the solenoid is $B = \mu_{0} \frac{N I}{l}$.The magnetic flux through each turn is $\phi = B A = \frac{\mu_{0} N I A}{l}$.The total flux linkage is $N \phi = \frac{\mu_{0} N^{2} I A}{l}$.The average induced back $emf$ is given by $e = \frac{\Delta(N \phi)}{\Delta t} = \frac{\mu_{0} N^{2} I A}{l \Delta t}$.Substituting the values:$e = \frac{4 \pi 	imes 10^{-7} 	imes (500)^{2} 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{4 	imes 3.14159 	imes 10^{-7} 	imes 250000 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{3.14159 	imes 10^{-7} 	imes 10^{6} 	imes 62.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e \approx 6.54 \; V \approx 6.5 \; V$.Thus,the average back $emf$ induced is $6.5 \; V$.

Similar Questions

The dimensions of self or mutual inductance are given as

The unit of inductance is

The self-inductance $L$ of a solenoid of length $\lambda$ and area of cross-section $A$ with a fixed number of turns $N$ increases as

$A$ graph of magnetic flux $(\phi)$ versus current $(I)$ is shown for four inductors $P, Q, R, S$. The largest value of self-inductance is for inductor:

$A$ solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid,the current will

Vedclass Test Series

Exam Paper Generator

Online Exam Module